I came across this question on the topology board at AoPS, where it hadn't really received an answer. It seems interesting, but I'm not sure how to solve it. Hopefully an answer will be found here.

The question, from Dudley's Real Analysis and Probability, goes as

A $C^1$ curve is a function $t\mapsto (f(t),g(t))$ from $\mathbb{R}$ into $\mathbb{R}^2$ where the derivatives $f'(t)$ and $g'(t)$ exist and are continuous for all $t$. Show that $\mathbb{R}^2$ is not a countable union of ranges of $C^1$ curves.

In the google book there's also a hint to show the range of a $C^1$ curve on a finite interval is nowhere dense.

So out of curiosity, what's a proper way to prove the hint that the range of a $C^1$ curve on a finite interval is nowhere dense? I've read through the first few sections of the book up to the point of the exercise, and it covers compactness, product topologies, complete and compact metric spaces and metrics on function spaces, and the theorems of Dini, Arzela-Ascoli, and the Stone Weierstrass Theorem. I'm putting a bounty to see if there's a basic solution using only this knowledge, even if it may not be the most elegant.

Added: I did a little research based on the comments I received. I know that $\mathbb{R}^2$ is a complete metric space, and the Baire Category Theorem says every complete metric space is of the second category. I can decompose a curve into countably many curves defined on finite intervals, (like intervals of length $1$ maybe?). Then if $\mathbb{R}^2$ is the countable union of ranges of curves, it would also be the countable union of the ranges of these countably many smaller curves with finite domains. However, if each curve on a finite interval has nowhere dense range, then $\mathbb{R}^2$ would be the union of countable many nowhere dense sets and then be of the first category, contrary to the Baire Category Theorem.

So why exactly is the range of a curve on a finite interval nowhere dense? It seems intuitively true. If I let $B=\{(f(t),g(t))\mid t\in I\}$ for some finite interval $I\subset\mathbb{R}$ be the range of some $C^1$ curve, then my feeling is this range looks like some "segment" in $\mathbb{R}^2$. If $U\subset\mathbb{R}^2$ is any open set, I want to find an open $V\subset U$ where $B\cap V=\emptyset$. Intuitively, it seems like I could just take some open ball of small enough radius in $U$ that doesn't meet $B$ in the plane, so the range is nowhere dense. Is there a more formal way to express this? Thanks.

The hint seems to be suggesting that you show the range of each curve is nowhere dense and then invoke the baire category theorem.
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JSchlatherOct 4 '11 at 23:58

3

You can "decompose" your curve to countably many curves defined on compact intervals. You will then get that $\mathbb{R}^2$ is a countable union of nowhere dense sets, aka a set of the first category. Use Baire's theorem to get a contradiction.
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MarkOct 4 '11 at 23:59

Thanks both of you. I read up on these things, and I think I have a better picture of what the question is suggesting now.
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yunoneOct 5 '11 at 2:15

yunone, the Baire category theorem appears before that exercise (as Theorem 2.5.2) and the hint clearly points to it. I would be surprised if there were a proof as you're asking for in the description of your bounty. So, I think the best you can hope for would be a detailed argument for the fact that the image of a $C^1$-curve is nowhere dense.
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t.b.Oct 8 '11 at 10:14

@t.b. Thanks for pointing this out, at this point I'm looking for a detailed argument that the image of a $C^1$ curve is nowhere dense. I'll try to make this more obvious in the body of the question.
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yunoneOct 8 '11 at 20:40

5 Answers
5

Here's an answer avoiding measure theory, or any advanced methods. First, if a curve $\gamma\colon\mathbb{R}\to\mathbb{R}^2$ is $C^1$ then, for any $a < b$, the restriction of $\gamma$ to $[a,b]$ has finite length $\int_a^b\vert\gamma^\prime(t)\vert\,dt$.

Now, a finite length curve cannot be dense the unit square $[0,1]^2$. For any positive integer $n$, consider the $(n+1)^2$ points $(i/n,j/n)$, $0\le i,j\le n$. The distance between any two of them is at least $1/n$. So, any curve which is dense in the unit square must join these points, and hence have length at least $((n+1)^2-1)/n=n+2$. Let $n$ go to infinity. By scaling, this shows that a finite length curve cannot be dense in any open subset of $\mathbb{R}^2$.

Now suppose that we have a countable set of $C^1$ curves $\gamma^i\colon\mathbb{R}\to\mathbb{R}^2$ ($i=1,2,\ldots$). Then, $\gamma^{ij}=\gamma^i\vert_{[-j,j]}$ ($i,j=1,2,\ldots$) is a countable set of curves of finite length, so their images are compact and nowhere dense subsets of $\mathbb{R}^2$.

Maybe I'm being dense, but I don't see how it follows immediately from the theorem you linked that the image of a $C^1$ curve has measure zero. I do agree it's a consequence, but the proof I have in mind requires at least a couple of observations. Do you have a completely trivial way of seeing this?
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Nick StrehlkeOct 5 '11 at 2:25

@Nick: the Jacobian of a $C^1$ function $\alpha : \mathbb{R} \to \mathbb{R}^2$ always has rank at most $1$, so every point is critical. Is the rest clear?
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Qiaochu YuanOct 5 '11 at 2:56

@Qiaochu Ah ok, I see what you were getting at. I somehow misunderstood the statement of the theorem when I read it, thinking that the hypothesis was $f:\mathbb R^n\to\mathbb R^m$ with $n\geq m$.
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Nick StrehlkeOct 5 '11 at 3:03

A $C^1$ curve on a compact interval has nowhere dense range: it is because it is Lipschitz, since its derivative is bounded. Lipschitz functions cannot increase the Hausdorff dimension, which is less than $2$ for an interval. So the range has two dimensional measure $0$, and it is closed, so it contains no open set.

You can construct a tubular neighborhood of such a curve having any desired area. It is the locus of points

$$(f(t),g(t)) + u \frac{\varepsilon(t)}{2}(g'(t),-f'(t))$$

where $t$ ranges over $\mathbb{R}$, $u$ ranges from $-1$ to $1$, and $\varepsilon(t)$ is suitably chosen. For instance, for all $t$ where the derivative of the curve is nonzero, let $\varepsilon(t) = |\varepsilon|/(\pi(1+t^2)||(f'(t),g'(t))||)$ (and otherwise let $\varepsilon(t)=0$) to drive the area down to $\varepsilon + O(\varepsilon^2)$, $\varepsilon \gt 0$. (Simple estimates will make this rigorous: it doesn't take any measure theory or fancy analysis. Just break the integral down into intervals over which neither $f'(t)$ nor $g'(t)$ varies much.)

Given a countable sequence of curves $(\gamma_i)_{1 \le i}$, create such tubular neighborhoods with $\varepsilon_i(t)$ chosen to make the tubular neighborhood of $\gamma_i$ less than $2^{-i}$ in area. The area of the union of those neighborhoods--which obviously includes the union of the curves--cannot exceed $\sum_i \varepsilon_i = 1$, whence it cannot fill $\mathbb{R}^2$.

This approach generalizes to differentiable submanifolds of $\mathbb{R}^n$.

If you want to do it via the Baire category theorem, here's a sketch of an approach. I'm not sure if it is the simplest, though.

We can reparametrize any curve so that the derivative $(f'(t), g'(t)) \ne (0,0)$ for all $t \in I$. [Actually, we can't necessarily do that.]

If $(f'(t), g'(t)) \ne (0,0)$, then there is an interval $[t-\epsilon, t+\epsilon]$ on which $(f(s), g(s))$ is injective.

We can cover $I$ by a finite union of such intervals.

Restricted to a set $[t-\epsilon, t+\epsilon]$ on which it is injective, the curve is an embedding: a homeomorphism onto its image. (It is an injective, continuous map from a compact space to a Hausdorff space.)

A homeomorphic copy of an interval in $\mathbb{R}^2$ must be nowhere dense. (It is closed, and if it contains a ball, then the unit interval contains a set homeomorphic to a ball of $\mathbb{R}^2$. Why is this impossible?)

The first step is not completely obvious to me and is what I was stuck on. Can you elaborate?
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Qiaochu YuanOct 5 '11 at 3:21

On reflection, it's not obvious because it is false. A curve with a corner or cusp can be the image of a $C^1$ function, and can't be reparametrized with nonzero derivative. Oops.
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Nate EldredgeOct 5 '11 at 12:51